Proving $\operatorname{Var}( E[ X \mid Z ] ) \le \operatorname{Var}( X )$

Question

I am trying to show that: $$\operatorname{Var} ( E[ X | Z ] ) \le \operatorname{Var}( X )$$ $X$ and $Z$ are random variables.

What I've tried $$\operatorname{Var}( X ) = E( X^2 )-(E[X])^2$$ $$\operatorname{Var}( E[ X | Z ] ) = E( E[X | Z]^2 )-(E[E[X|Z]])^2$$ According to the law of total expectation: $$\operatorname{Var}( E[ X | Z ] ) = E( E[X | Z]^2 )-([E[X])^2$$ Hence, I have to prove that: $$E( E[X | Z]^2 ) \le E( X^2 )$$ I read something about conditional Jensen inequality and that the expectation operator is an $L^2$ contraction but I don't understand this.

Thanks in advance!

Answer 1

Use the Law of Total Variance:

$ \operatorname{Var}(X) = E(\operatorname{Var}(X|Z)) + \operatorname{Var}(E(X|Z)) $

See: https://en.wikipedia.org/wiki/Law_of_total_variance

Answer 2

As you said, we need to prove:

$$E[E[X\mid Z]^2] \leq E [X^2]$$ Observe that $x\mapsto x^2$ is a convex map. Hence, by Jensen's inequality

$$E[X\mid Z ]^2 \le E[X^2 \mid Z]$$

Thus, because expectation is monotone:

$$E[E[X\mid Z]^2] \leq E[E[X^2\mid Z]] = E[X^2]$$

and the result follows.